The value of the integral integral_0^infinity (log_e (x))/(x^2+4) dx is:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q18 – JEE-MAIN 2026

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Home Maths Limits and DerivativesQuestion #263

The value of the integral

\int_{0}^{\infty} \frac{\log_{e} (x)}{x^{2} + 4} d x

is:

Clue 1 of 3: Feynman's Technique (Differentiation Under the Integral Sign)

Consider a generalized integral $I (s) = \int_{0}^{\infty} \frac{x^{s}}{x^{2} + 4} d x$ and relate the given integral to its derivative with respect to $s$ at $s = 0$ .

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Step 1: Generalize the Integral✦ Active

Define a generalized integral $I (s) = \int_{0}^{\infty} \frac{x^{s}}{x^{2} + 4} d x$ . The given integral is $I = \int_{0}^{\infty} \frac{\log x}{x^{2} + 4} d x$ . We can relate the given integral to the derivative of $I (s)$ with respect to $s$ evaluated at $s = 0$ :

I = {\frac{d}{d s} (\int_{0}^{\infty} \frac{x^{s}}{x^{2} + 4} d x) |}_{s = 0} = {\frac{d I (s)}{d s} |}_{s = 0}

Step 2: Evaluate the Generalized Integral

I (s)

○ Expand

Perform a substitution $x = 2 t$ , which implies $d x = 2 d t$ . The limits of integration remain from $0$ to $\infty$ .

I (s) = \int_{0}^{\infty} \frac{(2 t)^{s}}{(2 t)^{2} + 4} 2 d t = \int_{0}^{\infty} \frac{2^{s} t^{s}}{4 t^{2} + 4} 2 d t = \frac{2^{s + 1}}{4} \int_{0}^{\infty} \frac{t^{s}}{t^{2} + 1} d t = 2^{s - 1} \int_{0}^{\infty} \frac{t^{s}}{t^{2} + 1} d t

Using the known result for the integral $\int_{0}^{\infty} \frac{t^{s}}{t^{2} + 1} d t = \frac{π}{2 \cos (\frac{π s}{2})}$ for $- 1 < Re (s) < 1$ , we get:

I (s) = 2^{s - 1} \frac{π}{2 \cos (\frac{π s}{2})}

💡 Teacher's Secret Hint

Remember the conditions for the known integral formula.

Step 3: Differentiate and Evaluate at

s = 0

○ Expand

Now, differentiate $I (s)$ with respect to $s$ using the quotient rule $\frac{d}{d s} (\frac{u}{v}) = \frac{u^{'} v - u v^{'}}{v^{2}}$ , where $u = 2^{s - 1}$ and $v = \cos (\frac{π s}{2})$ . Then, substitute $s = 0$ .

\frac{d I (s)}{d s} = \frac{π}{2} \frac{(2^{s - 1} \log 2) \cos (\frac{π s}{2}) - 2^{s - 1} (- \sin (\frac{π s}{2}) \frac{π}{2})}{\cos^{2} (\frac{π s}{2})}

Substitute $s = 0$ : $2^{0 - 1} = \frac{1}{2}$ , $\cos (0) = 1$ , $\sin (0) = 0$ .

{\frac{d I (s)}{d s} |}_{s = 0} = \frac{π}{2} \frac{(\frac{1}{2} \log 2) \cdot 1 - \frac{1}{2} (0)}{1^{2}} = \frac{π}{2} \cdot \frac{1}{2} \log 2 = \frac{π \log 2}{4}

Thus, the value of the integral is $\frac{π \log_{e} (2)}{4}$ .

💡 Teacher's Secret Hint

Be careful with the chain rule when differentiating trigonometric functions.

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Maths Question 18 – JEE-MAIN 2026

Clue 1 of 3: Feynman's Technique (Differentiation Under the Integral Sign)

🧠 Full Solution Path